Difference between revisions of "2018 AMC 8 Problems/Problem 24"
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Note that <math>EJCI</math> is a rhombus. | Note that <math>EJCI</math> is a rhombus. | ||
Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= \sqrt 3s</math> and <math>JI=\sqrt 2s</math>. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is <math>\frac{\sqrt 6s^2}{2}</math>. <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math> | Let the side length of the cube be <math>s</math>. By the Pythagorean theorem, <math>EC= \sqrt 3s</math> and <math>JI=\sqrt 2s</math>. Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is <math>\frac{\sqrt 6s^2}{2}</math>. <math>R = \frac{\sqrt 6}2</math>. Thus <math>R^2 = \boxed{\textbf{(C) } \frac{3}{2}}</math> | ||
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+ | ==Note== | ||
+ | In the 2008 AMC 10A, Question 21 was nearly identical to this question. | ||
==See Also== | ==See Also== |
Revision as of 11:35, 22 November 2018
Contents
Problem 24
In the cube with opposite vertices and and are the midpoints of edges and respectively. Let be the ratio of the area of the cross-section to the area of one of the faces of the cube. What is
Solution
Note that is a rhombus. Let the side length of the cube be . By the Pythagorean theorem, and . Since the area of a rhombus is half the product of it's diagonals, so the area of the cross section is . . Thus
Note
In the 2008 AMC 10A, Question 21 was nearly identical to this question.
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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